The equation $y = -4.9t^2 + 23.8t$ describes the height (in meters) of a projectile launched from the ground at 23.8 meters per second. In how many seconds will the projectile first reach 28 meters in height?
Solution: Setting $y$ to 28, we find the following: \begin{align*}
28& = -4.9t^2 + 23.8t\\
0 & = -4.9t^2 + 23.8t - 28\\
0 & = 49t^2 - 238t + 280\\
& = 7t^2 - 34t + 40\\
& = (7t - 20)(t - 2)
\end{align*}Our possible values for $t$ are $\frac{20}{7} \approx 2.857$ or $2.$ Of these, we choose the smaller $t$, or $\boxed{2}.$